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A, B and C Can Reap a Field in 15 3 4 Days; B, C and D in 14 Days; C, D and a in 18 Days; D, a and B in 21 Days. in What Time Can A, B, C and D Together Reap It? - Mathematics

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ConceptTime and Work

Question

AB and C can reap a field in \[15\frac{3}{4}\] days; BC and D in 14 days; CD and A in 18 days; DA and B in 21 days. In what time can ABC and D together reap it?

 

Solution

\[\text{ Time taken by } \left( A + B + C \right) \text{ to do the work } = 15\frac{3}{4} \text{ days } = \frac{63}{4} \text{ days } \]
\[\text{ Time taken by }  \left( B + C + D \right) \text{ to do the work = 14 days} \]
\[\text{ Time taken by }  \left( C + D + A \right) \text{ to do the work = 18 days } \]
\[\text{ Time taken by } \left( D + A + B \right) \text{ to do the work = 21 days } \]
\[\text{ Now, }  \]
\[ \text{ Work done by } \left( A + B + C \right) = \frac{4}{63}\]
\[\text{ Work done by }  \left( B + C + D \right) = \frac{1}{14}\]
\[ \text{ Work done by } \left( C + D + A \right) = \frac{1}{18}\]
\[ \text{ Work done by } \left( D + A + B \right) = \frac{1}{21}\]
\[ \therefore \text{ Work done by working together }  = \left( A + B + C \right) + \left( B + C + D \right) + \left( C + A + D \right) + \left( D + A + B \right)\]
\[ = \frac{4}{63} + \frac{1}{14} + \frac{1}{18} + \frac{1}{21}\]
\[ = \frac{4}{63} + \left( \frac{9 + 7 + 6}{126} \right) = \frac{4}{63} + \frac{22}{126}\]
\[ = \frac{4}{63} + \frac{11}{63} = \frac{15}{63}\]
\[ \therefore \text{ Work done by working together }  = 3\left( A + B + C + D \right) = \frac{15}{63}\]
\[ \therefore \text{ Work done by } \left( A + B + C + D \right) = \frac{15}{63 \times 3} = \frac{5}{63}\]
\[ \text{ Thus, together they can do the work in } \frac{63}{5} \text{ days or }  12\frac{3}{5} \text{ days }  .\]

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Solution A, B and C Can Reap a Field in 15 3 4 Days; B, C and D in 14 Days; C, D and a in 18 Days; D, a and B in 21 Days. in What Time Can A, B, C and D Together Reap It? Concept: Time and Work.
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